mystery3 method
public static int sumEvenDigits(int n)
{
if(n == 0)
return 0;
if(n % 2 == 0)
return n % 10 + sumEvenDigits(n / 10);
else
return sumEvenDigits(n / 10);
}
The method returns the sum of the even digits of its parameter n.
The analysis is nearly identical to mystery1/countOddDigits. The only difference is that on 1 step, the method adds n % 10 (the last digit of n) to the result, instead of adding 1.